A168538 a(n) = (n+6)*(n+1)*(n^2 + 7*n + 16)/4.

0, 24, 84, 204, 414, 750, 1254, 1974, 2964, 4284, 6000, 8184, 10914, 14274, 18354, 23250, 29064, 35904, 43884, 53124, 63750, 75894, 89694, 105294, 122844, 142500, 164424, 188784, 215754, 245514, 278250, 314154, 353424, 396264, 442884, 493500, 548334

Offset

-1,2

Comments

In general, a sequence of the form a(n) = sum(k+x+3)!/(k+x)!,k=1..n) will have a closed form of n/4 * (n + 5 + 2x)*(n^2 + 5n + 2xn + 10 + 2x^2 + 10x). - Gary Detlefs, Aug 10 2010

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

From R. J. Mathar, Mar 21 2010: (Start)

a(n) = 6*A063258(n).

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

G.f.: 6*(x-2)*(x^2 - 2*x + 2)/(x-1)^5. (End)

From Gary Detlefs, Aug 10 2010: (Start)

a(n) = sum_{k=1..n} (k+3)*(k+2)*(k+1), with offset 0.

a(n) = (n/4)*(n+5)*(n^2 + 5n + 10), with offset 0. (End)

E.g.f.: (1/4)*(96 + 240*x + 120*x^2 + 20*x^3 + x^4)*exp(x). - G. C. Greubel, Jul 25 2016

Mathematica

LinearRecurrence[{5, -10, 10, -5, 1}, {0, 24, 84, 204, 414}, 25] (* G. C. Greubel, Jul 25 2016 *)
Table[1/4 (n+6)(n+1)(n^2+7n+16), {n, -1, 40}] (* Harvey P. Dale, Jul 07 2019 *)

Prog

(PARI) a(n) = (n+6)*(n+1)*(n^2 + 7*n + 16)/4; \\ Michel Marcus, Jan 10 2015
(Magma) [(n+6)*(n+1)*(n^2+7*n+16)/4: n in [-1..35]]; // Vincenzo Librandi, Jul 26 2016

Crossrefs

Sequence in context: A289155 A101861 A280304 * A007201 A228874 A211328

Adjacent sequences: A168535 A168536 A168537 * A168539 A168540 A168541

Keyword

nonn,easy

Author

Kristo Jorgenson (kristoj(AT)me.com), Nov 29 2009

Extensions

Edited by N. J. A. Sloane, Nov 29 2009

Status

approved